Question

A test rocket is fired straight up from rest. The net acceleration is 20 m/s2 upward and continues for 4.0 seconds, at which time the rocket engines cease firing. What maximum elevation does the rocket reach?

Answer 1

486.5 m

Step-by-step explanation:

Initial velocity is zero as the rocket is fired from rest. u = 0.

Displacement of the rocket during this time:

s = ut +0.5 at²

s = 0+0.5 ×20×4²

s = 160 m

The final velocity at the end of 4 s is:

v = u + at

v = 0 + (20)(4)

v = 80 m/s

This will become the initial velocity for the next half of the motion.

At the maximum elevation, velocity is zero. v = 0

Acceleration due to gravity always acts downwards.

`s=(v^2-u^2)/(2a)\\s=(0-80^2)/(2* -9.8) = 326.5 m`

Thus, the maximum elevation that test rocket would reach is:

326.5 m + 160 m = 486.5 m

Answer 2

Maximum elevation, h = 160 meters

Step-by-step explanation:

Initially, the rocket is at rest, u = 0

Acceleration of the rocket,
`a=20\ m/s^2`

Time, t = 4 s

We need to find the maximum elevation reached by the rocket. It can be calculated using second equation of motion as :

`h=ut+(1)/(2)* a* t^2`

`h=(1)/(2)* 20* (4)^2`

h = 160 meters

So, the maximum elevation reached by the rocket is 160 meters. Hence, this is the required solution.